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mr Goodwill [35]
3 years ago
11

What is the distance ?

Mathematics
1 answer:
aleksklad [387]3 years ago
8 0

The solution is you need to subtract the y axis and x axis

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Please helpppp!!! Find the area of the irregular figure
OverLord2011 [107]

Answer:

The area of the figure is 143 sq in

Step-by-step explanation:

10 in * 7 in = 70 sq in

12 in * 4 in = 48 sq in

5 in * 5 in = 25 sq in

A = 70 sq in + 48 sq in + 25 sq in

A = 143 sq in

3 0
2 years ago
What happened that caused Brian to lose his shelter and all of his belongings? (picture in link) helpp pls fast!
Lilit [14]

Answer:

Step-by-step explanation:

Can I have the rest of the question(picture)?

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3 years ago
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Please help! asap! i have no idea what it is and it’s worth a lot of points on my test!
beks73 [17]
I haven’t done math in forever, do you want us to simplify it? Bc it would be 2x+8=-16=4x+6 but I’m not sure
5 0
2 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
Triangle congruence... SAS SSS AAS ASA pls explain why
abruzzese [7]

Answer:

it is SAS

Step-by-step explanation:

because BC is common side in both triangles

so

now two triangles have 2 sides equal and 1 angle equal so

it is congruent by SAS rule

4 0
2 years ago
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